In the production of aluminum, 2 tons of alumina AI2O3, which is an integral part of clay, are consumed

In the production of aluminum, 2 tons of alumina AI2O3, which is an integral part of clay, are consumed for each ton of aluminum. Calculate the mass fraction (%) of the output of aluminum from the theoretically possible.

Given:
m pract. (Al) = 1 t = 1,000,000 g
m (Al2O3) = 2 t = 2,000,000 g

To find:
η (Al) -?

1) 2Al2O3 = (electric current) => 4Al + 3O2;
2) n (Al2O3) = m / M = 2,000,000 / 102 = 19607.8 mol;
3) n theory. (Al) = n (Al2O3) * 2 = 19607.8 * 2 = 39215.6 mol;
4) m theor. (Al) = n theory. * M = 39215.6 * 27 = 1058821.2 g;
5) η (Al) = m practical. * 100% / m theory. = 1,000,000 * 100% / 1,058,821.2 = 94%.

Answer: The Al yield is 94%.